Given a set S (all distinct elements) of integers, find the largest d such that a + b + c = d
where a, b, c, and d are distinct elements of S.
Constraints: 1 ? number of elements in the set ? 1000 INT_MIN ? each element in the set ? INT_MAX
Examples :
Input : S[] = {2, 3, 5, 7, 12}
Output : 12
Explanation: 12 is the largest d which can be represented as 12 = 2 + 3 + 7Input : S[] = {2, 16, 64, 256, 1024}
Output : No solution
Method 1(Brute Force): We can solve this problem using simple brute force approach which is not very efficient as |S| can be as large as 1000. We’ll sort the set of elements and start by finding the largest d by equating it with the sum of all possible combinations of a, b and c.
Steps to solve the problem:
- Sort the input array S in ascending order.
- Iterate the input array S in reverse order. For each S[i]:
- Iterate over all unique combinations of three elements j, k, l from the remaining elements in S, such that j < k < l and i != j != k !=
- if S[j] + S[k] + S[l] == S[i], return S[i] as the largest possible value for d.
- If no such combination is found, return INT_MIN.
Below is the implementation of above idea :
C++
// CPP Program to find the largest d// such that d = a + b + c#include <bits/stdc++.h>using namespace std;int findLargestd(int S[], int n){ bool found = false; // sort the array in // ascending order sort(S, S + n); // iterating from backwards to // find the required largest d for (int i = n - 1; i >= 0; i--) { for (int j = 0; j < n; j++) { // since all four a, b, c, // d should be distinct if (i == j) continue; for (int k = j + 1; k < n; k++) { if (i == k) continue; for (int l = k + 1; l < n; l++) { if (i == l) continue; // if the current combination // of j, k, l in the set is // equal to S[i] return this // value as this would be the // largest d since we are // iterating in descending order if (S[i] == S[j] + S[k] + S[l]) { found = true; return S[i]; } } } } } if (found == false) return INT_MIN;}// Driver Codeint main(){ // Set of distinct Integers int S[] = { 2, 3, 5, 7, 12 }; int n = sizeof(S) / sizeof(S[0]); int ans = findLargestd(S, n); if (ans == INT_MIN) cout << "No Solution" << endl; else cout << "Largest d such that a + b + " << "c = d is " << ans << endl; return 0;} |
Java
// Java Program to find the largest// such that d = a + b + cimport java.io.*;import java.util.Arrays;class GFG { // function to find largest d static int findLargestd(int []S, int n){ boolean found = false; // sort the array in // ascending order Arrays.sort(S); // iterating from backwards to // find the required largest d for (int i = n - 1; i >= 0; i--) { for (int j = 0; j < n; j++) { // since all four a, b, c, // d should be distinct if (i == j) continue; for (int k = j + 1; k < n; k++) { if (i == k) continue; for (int l = k + 1; l < n; l++) { if (i == l) continue; // if the current combination // of j, k, l in the set is // equal to S[i] return this // value as this would be the // largest d since we are // iterating in descending order if (S[i] == S[j] + S[k] + S[l]) { found = true; return S[i]; } } } } } if (found == false) return Integer.MAX_VALUE; return -1;}// Driver Codepublic static void main(String []args) { // Set of distinct Integers int []S = new int[]{ 2, 3, 5, 7, 12 }; int n = S.length; int ans = findLargestd(S, n); if (ans == Integer.MAX_VALUE) System.out.println("No Solution"); else System.out.println("Largest d such that " + "a + " + "b + c = d is " + ans ); } }// This code is contributed by Sam007 |
Python3
# Python Program to find the largest # d such that d = a + b + cdef findLargestd(S, n) : found = False # sort the array in ascending order S.sort() # iterating from backwards to # find the required largest d for i in range(n-1, -1, -1) : for j in range(0, n) : # since all four a, b, c, # d should be distinct if (i == j) : continue for k in range(j + 1, n) : if (i == k) : continue for l in range(k+1, n) : if (i == l) : continue # if the current combination # of j, k, l in the set is # equal to S[i] return this # value as this would be the # largest d since we are # iterating in descending order if (S[i] == S[j] + S[k] + S[l]) : found = True return S[i] if (found == False) : return -1# Driver Code# Set of distinct IntegersS = [ 2, 3, 5, 7, 12 ]n = len(S)ans = findLargestd(S, n)if (ans == -1) : print ("No Solution") else : print ("Largest d such that a + b +" , "c = d is" ,ans)# This code is contributed by Manish Shaw# (manishshaw1) |
C#
// C# Program to find the largest// such that d = a + b + cusing System;class GFG { // function to find largest d static int findLargestd(int []S, int n){ bool found = false; // sort the array // in ascending order Array.Sort(S); // iterating from backwards to // find the required largest d for (int i = n - 1; i >= 0; i--) { for (int j = 0; j < n; j++) { // since all four a, b, c, // d should be distinct if (i == j) continue; for (int k = j + 1; k < n; k++) { if (i == k) continue; for (int l = k + 1; l < n; l++) { if (i == l) continue; // if the current combination // of j, k, l in the set is // equal to S[i] return this // value as this would be the // largest dsince we are // iterating in descending order if (S[i] == S[j] + S[k] + S[l]) { found = true; return S[i]; } } } } } if (found == false) return int.MaxValue; return -1;}// Driver Codepublic static void Main() { // Set of distinct Integers int []S = new int[]{ 2, 3, 5, 7, 12 }; int n = S.Length; int ans = findLargestd(S, n); if (ans == int.MaxValue) Console.WriteLine( "No Solution"); else Console.Write("Largest d such that a + " + "b + c = d is " + ans ); } }// This code is contributed by Sam007 |
PHP
<?php// PHP Program to find the largest // d such that d = a + b + cfunction findLargestd( $S, $n){$found = false; // sort the array in // ascending order sort($S); // iterating from backwards to // find the required largest d for ( $i = $n - 1; $i >= 0; $i--) { for ( $j = 0; $j < $n; $j++) { // since all four a, b, c, // d should be distinct if ($i == $j) continue; for ( $k = $j + 1; $k < $n; $k++) { if ($i == $k) continue; for ( $l = $k + 1; $l < $n; $l++) { if ($i == $l) continue; // if the current combination // of j, k, l in the set is // equal to S[i] return this // value as this would be the // largest d since we are // iterating in descending order if ($S[$i] == $S[$j] + $S[$k] + $S[$l]) { $found = true; return $S[$i]; } } } } } if ($found == false) return PHP_INT_MIN;}// Driver Code// Set of distinct Integers$S = array( 2, 3, 5, 7, 12 );$n = count($S);$ans = findLargestd($S, $n);if ($ans == PHP_INT_MIN) echo "No Solution" ;else echo "Largest d such that a + b + " , "c = d is " , $ans ;// This code is contributed by anuj_67.?> |
Javascript
<script> // Javascript Program to find the largest // such that d = a + b + c // function to find largest d function findLargestd(S, n) { let found = false; // sort the array // in ascending order S.sort(); // iterating from backwards to // find the required largest d for (let i = n - 1; i >= 0; i--) { for (let j = 0; j < n; j++) { // since all four a, b, c, // d should be distinct if (i == j) continue; for (let k = j + 1; k < n; k++) { if (i == k) continue; for (let l = k + 1; l < n; l++) { if (i == l) continue; // if the current combination // of j, k, l in the set is // equal to S[i] return this // value as this would be the // largest dsince we are // iterating in descending order if (S[i] == S[j] + S[k] + S[l]) { found = true; return S[i]; } } } } } if (found == false) return Number.MAX_VALUE; return -1; } // Set of distinct Integers let S = [ 2, 3, 5, 7, 12 ]; let n = S.length; let ans = findLargestd(S, n); if (ans == Number.MAX_VALUE) document.write( "No Solution"); else document.write("Largest d such that a + " + "b + c = d is " + ans ); </script> |
Output :
Largest d such that a + b + c = d is 12
This brute force solution has a time complexity of O((size of Set)4).
Method 2(Efficient Approach – Using Hashing): The above problem statement (a + b + c = d) can be restated as finding a, b, c, d such that a + b = d – c. So this problem can be efficiently solved using hashing.
- Store sums of all pairs (a + b) in a hash table
- Traverse through all pairs (c, d) again and search for (d – c) in the hash table.
- If a pair is found with the required sum, then make sure that all elements are distinct array elements and an element is not considered more than once.
Below is the implementation of above approach.
C++
// A hashing based CPP program to find largest d// such that a + b + c = d.#include <bits/stdc++.h>using namespace std;// The function finds four elements with given sum Xint findFourElements(int arr[], int n){ // Store sums (a+b) of all pairs (a,b) in a // hash table unordered_map<int, pair<int, int> > mp; for (int i = 0; i < n - 1; i++) for (int j = i + 1; j < n; j++) mp[arr[i] + arr[j]] = { i, j }; // Traverse through all pairs and find (d -c) // is present in hash table int d = INT_MIN; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int abs_diff = abs(arr[i] - arr[j]); // If d - c is present in hash table, if (mp.find(abs_diff) != mp.end()) { // Making sure that all elements are // distinct array elements and an element // is not considered more than once. pair<int, int> p = mp[abs_diff]; if (p.first != i && p.first != j && p.second != i && p.second != j) d = max(d, max(arr[i], arr[j])); } } } return d;}// Driver program to test above functionint main(){ int arr[] = { 2, 3, 5, 7, 12 }; int n = sizeof(arr) / sizeof(arr[0]); int res = findFourElements(arr, n); if (res == INT_MIN) cout << "No Solution."; else cout << res; return 0;} |
Java
// A hashing based Java program to find largest d // such that a + b + c = d. import java.util.HashMap;import java.lang.Math;// To store and retrieve indices pair i & jclass Indexes { int i, j; Indexes(int i, int j) { this.i = i; this.j = j; } int getI() { return i; } int getJ() { return j; }}class GFG { // The function finds four elements with given sum X static int findFourElements(int[] arr, int n) { HashMap<Integer, Indexes> map = new HashMap<>(); // Store sums (a+b) of all pairs (a,b) in a // hash table for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { map.put(arr[i] + arr[j], new Indexes(i, j)); } } int d = Integer.MIN_VALUE; // Traverse through all pairs and find (d -c) // is present in hash table for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int abs_diff = Math.abs(arr[i] - arr[j]); // If d - c is present in hash table, if (map.containsKey(abs_diff)) { Indexes indexes = map.get(abs_diff); // Making sure that all elements are // distinct array elements and an element // is not considered more than once. if (indexes.getI() != i && indexes.getI() != j && indexes.getJ() != i && indexes.getJ() != j) { d = Math.max(d, Math.max(arr[i], arr[j])); } } } } return d; } // Driver code public static void main(String[] args) { int arr[] = { 2, 3, 5, 7, 12 }; int n = arr.length; int res = findFourElements(arr, n); if (res == Integer.MIN_VALUE) System.out.println("No Solution"); else System.out.println(res); }}// This code is contributed by Vivekkumar Singh |
Python3
# A hashing based Python3 program to find # largest d, such that a + b + c = d.# The function finds four elements# with given sum Xdef findFourElements(arr, n): # Store sums (a+b) of all pairs (a,b) in a # hash table mp = dict() for i in range(n - 1): for j in range(i + 1, n): mp[arr[i] + arr[j]] =(i, j) # Traverse through all pairs and find (d -c) # is present in hash table d = -10**9 for i in range(n - 1): for j in range(i + 1, n): abs_diff = abs(arr[i] - arr[j]) # If d - c is present in hash table, if abs_diff in mp.keys(): # Making sure that all elements are # distinct array elements and an element # is not considered more than once. p = mp[abs_diff] if (p[0] != i and p[0] != j and p[1] != i and p[1] != j): d = max(d, max(arr[i], arr[j])) return d# Driver Codearr = [2, 3, 5, 7, 12]n = len(arr)res = findFourElements(arr, n)if (res == -10**9): print("No Solution.")else: print(res)# This code is contributed by Mohit Kumar |
C#
// A hashing based C# program to find // largest d such that a + b + c = d. using System;using System.Collections.Generic;// To store and retrieve // indices pair i & jpublic class Indexes { int i, j; public Indexes(int i, int j) { this.i = i; this.j = j; } public int getI() { return i; } public int getJ() { return j; }}public class GFG{ // The function finds four elements // with given sum X static int findFourElements(int[] arr, int n) { Dictionary<int, Indexes> map = new Dictionary<int, Indexes>(); // Store sums (a+b) of all pairs // (a,b) in a hash table for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { map.Add(arr[i] + arr[j], new Indexes(i, j)); } } int d = int.MinValue; // Traverse through all pairs and // find (d -c) is present in hash table for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int abs_diff = Math.Abs(arr[i] - arr[j]); // If d - c is present in hash table, if (map.ContainsKey(abs_diff)) { Indexes indexes = map[abs_diff]; // Making sure that all elements are // distinct array elements and an element // is not considered more than once. if (indexes.getI() != i && indexes.getI() != j && indexes.getJ() != i && indexes.getJ() != j) { d = Math.Max(d, Math.Max(arr[i], arr[j])); } } } } return d; } // Driver code public static void Main(String[] args) { int []arr = { 2, 3, 5, 7, 12 }; int n = arr.Length; int res = findFourElements(arr, n); if (res == int.MinValue) Console.WriteLine("No Solution"); else Console.WriteLine(res); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// A hashing based Javascript program to find largest d// such that a + b + c = d.// To store and retrieve indices pair i & jclass Indexes{ constructor(i,j) { this.i = i; this.j = j; } getI() { return this.i; } getJ() { return this.j; }}// The function finds four elements with given sum Xfunction findFourElements(arr,n){ let map = new Map(); // Store sums (a+b) of all pairs (a,b) in a // hash table for (let i = 0; i < n - 1; i++) { for (let j = i + 1; j < n; j++) { map.set(arr[i] + arr[j], new Indexes(i, j)); } } let d = Number.MIN_VALUE; // Traverse through all pairs and find (d -c) // is present in hash table for (let i = 0; i < n - 1; i++) { for (let j = i + 1; j < n; j++) { let abs_diff = Math.abs(arr[i] - arr[j]); // If d - c is present in hash table, if (map.has(abs_diff)) { let indexes = map.get(abs_diff); // Making sure that all elements are // distinct array elements and an element // is not considered more than once. if (indexes.getI() != i && indexes.getI() != j && indexes.getJ() != i && indexes.getJ() != j) { d = Math.max(d, Math.max(arr[i], arr[j])); } } } } return d;}// Driver codelet arr=[ 2, 3, 5, 7, 12];let n = arr.length;let res = findFourElements(arr, n);if (res == Number.MIN_VALUE) document.write("No Solution");else document.write(res);// This code is contributed by patel2127</script> |
Output:
12
The overall Time Complexity for this efficient approach is O(N2) (where N is the size of the set).
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